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Studying for a test? Prepare with these 5 lessons on Alkanes, cycloalkanes, and functional groups.

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# Bicyclic compounds

Video transcript

In the last few
videos, we looked at the different
conformations of cyclohexane. And obviously, cyclohexane
is just one ring, so it'd be a
monocyclic compound. The official IUPAC way of
telling how many rings you have is how many cuts does it take
to make an open chain alkane? So if I just made a cut
between those two carbons, I have an open chain alkane. And it took me one
cut to get there. So obviously, cyclohexane
is monocyclic. What about if I had
a cyclohexane ring, and then I had another
ring fused to it like that? Obviously, there
are two rings here, so this would be a
bicyclic compound. Once again, if you were to use
the IUPAC way of figuring out how many rings you
have, you would make a cut here and a
cut here, and now you have an open chain alkane. So it took you two
cuts to get there, so you would classify this
as a bicyclic molecule. Now, these are very
easy, simple compounds. But for more
complicated compounds, you'll need to remember that
IUPAC rule to figure out how many rings
you actually have. Let's go ahead and redraw
that second molecule here. And let's look at it
in some more detail. So I had a cyclohexane
ring, and then I had another ring
fused to it like that. Now, the carbons that
are shared by both of these rings, so this
carbon and this carbon, those are called your
bridgehead carbons. Let's go ahead and write that. The bridgehead carbons. And when you're numbering
your bicyclic compounds, you want to start at the
bridgehead carbon, then number along the longest path,
then the second longest path, and then finally
the shortest path. So let's go ahead and do that. Let's start with the
top bridgehead carbon. We'll make that number 1. And we're going to number
along the longest path, so the longest path
would be going left. So that's 1, 2, 3, 4, 5, and 6. And now, the second
longest path. Well, we can just continue,
and there'd be 7 and 8 since that's the
second longest path. The third path,
the shortest path, is usually the one between
your bridgehead carbons. And there are no carbons
between our bridgehead carbons. So we're done in terms of
numbering our compound. The general formula for naming
a bicyclo compound-- we've already determined this
is a bicyclo compound, so you would have bicyclo here. And then in brackets, you
would put the number of carbons in the longest path excluding
the bridgehead carbon, which I will represent here with x. And then y is the
number of carbons of the second longest
path, excluding the bridgehead carbon. And then z is the number of
carbons in the shortest path, excluding the bridgehead carbon. And then you finish everything
off with the alkane. So how many carbons are
there in this molecule? So for this example, let's
go ahead and name it. We've already determined that
it's a bicyclic compound, so we'll go ahead and
write bicyclo here. And x, again, is the number of
carbons in the longest path, excluding the bridgehead carbon. So we're going to find
the longest path, which is on the left here,
and we're going to ignore the bridgehead carbon. How many carbons where there? 1, 2, 3, and 4. So we'll go ahead
and put a 4 here. y was the number of carbons
in the second longest path. Well, that was
over here where we had this carbon and this carbon. Again, excluding the
bridgehead carbon. So that would get a 2. And finally, the
shortest path, which was how many carbons are there
between my bridgehead carbons. And there are, of
course, 0 carbons between my bridgehead carbons. So if there's 0 carbons between
your bridgehead carbons, obviously your
bridgehead carbons are bonded directly
together, which we can see in the dot structure. Finally, how many
carbons are there total? There are 8, and we
know that means octane. So the final name
for this molecule is bicyclo[4.2.0.]octane. Let's do another one following
our general formula here for naming a bicyclo compound. So let's take a
cyclohexane ring. And let's make this
a bridgehead carbon. And we're actually
going to put a carbon between our bridgehead
carbons like that. Now, this is hard to see
when it's drawn this way. So usually, you'll see it
drawn a little bit differently. Usually you'll see it
drawn with a little bit more three-dimensionality to it. So hopefully we
can do that here. So it looks something like this. So let's find our
bridgehead carbons here. So we know that this
is a bridgehead carbon, this is a bridgehead carbon. Those correspond to
these guys over here. And how many cuts would
it take to turn this into an open chain alkane? Well, let's go ahead and look
at the drawing on the left. And if we cut it right
here, and we cut it here, we can see it's now an
open chain compound. So it took two cuts,
so it's bicyclo. Let's go ahead and put those
bonds back in there like that. So it's a bicyclo compound. So we can go ahead
and start naming it as a bicyclo compound. We need to go ahead
and number it, so let's start with
our bridgehead carbon. So we'll start with this
is our bridgehead carbon. Find the longest path. Well, it doesn't matter if
I go left or right here, since it's the same
length each way around. So I'll just go to the right. So 1, 2, 3, 4. Second longest
path, so I'm going to keep on going
this way, so 5 and 6. And then finally,
the shortest path, which is the one carbon
in between my two bridgehead carbons,
which would then get a 7, so seven total carbons. All right, so when
I'm naming this, I need to put my
brackets in here. And I put the number of
carbons in the longest path, excluding the
bridgehead carbon here. So how many carbons were
my longest path, excluding the bridgehead? Here's 1 and here is 2, so I'm
going to go ahead and put a 2 in here. And then I go to my
second longest path. Well, that was
over on this side. And there are also two carbons
on my second longest path. Obvious, it's the same
length as the other one. So it's 2, 2. And then my shortest path,
how many carbons are there in my shortest path, not
counting my bridgehead carbon? There's one. So I go ahead and put a 1 here. And so I have bicyclo[2.2.1]. And then my total
number of carbons is 7. So this is a
bicycloheptane molecules. So bicyclobicyclo[2.2.1]heptane would
be the official IUPAC name for this molecule. This molecule turns
up a lot in nature. So much so, that it actually
has its own common name. This is also called norborane. So let's go ahead
and write that here. So this is norborane,
again, a structure found often in nature. All right, let's do
one more example here. And let's make it look similar
to that norborane molecule. So we'll have it go like this. So it definitely is
a little bit tricky to draw these molecules here. So let's put some
substituents on this molecule. So I'm going to attempt to draw
the same molecule over here. And we're going to take a look
at two different ways to number it and see which one
turns out to be correct. So here I have the exact same
molecule on both sides here. And let's see how to do this. So let's first find
our bridgehead carbons. So here are my bridgehead
carbons right here. Now, when I start
numbering, I could start at either one of
those bridgehead carbons. So if I identify my bridgehead
carbons over on this one, I could start numbering with
either of these be number one. So let's just make
this number 1 here. So if that's number 1, I
next go to my longest path. So that would be going
to the right here. So this would give me a
2 for this carbon, a 3 for this carbon, a 4 for this
carbon, a 5 for this carbon. Next, the second longest path. Well, I can just keep going,
make that a 6, make this a 7. And then finally,
my shortest path, which is up here at the top. So that would be 8 total
carbons for my parent name. On the dot structure
on the left, again, the exact
same dot structure. This time, I'm going to start
with the opposite bridgehead carbons. So I'm going to start with
this bridgehead carbon. And then follow the same rule. So the longest path, which
would be to the right here. So 2, 3, 4, 5, 6, 7, and then 8. So which one of these numbering
systems is the correct one? Remember, our goal
is to get the lowest number possible for
our substituents. And if I look at the
example on the right, I have a methyl group
in the 7 position and a methyl group
in the 8 position. The example on the left has a
methyl group in the 6 position and a methyl group
in the 8 position. Since I want to give
my lowest number possible to my substituent, I'm
going to not number it the way on the right. So I'm going to go with
the way on the left here. So let's go ahead
and start naming it. We already know this
is a bicyclo compound. And we have two methyl
groups, one at 6 and one at 8. So we could start naming
it by saying 6,8-dimethyl. And then we know this is a
bicyclo compound, so bicyclo. And then in the
brackets, we're going to do the longest
path first, excluding the bridgehead carbon. So the longest path
is on the right. How many carbons are
there in the longest path? There are three excluding
the bridgehead carbon, so we'll go ahead
and put a 3 in here. Next, the second
longest path, excluding the bridgehead carbon. Well, that was these
two right here. So next, there'll be a 2. And then finally,
the shortest path excluding our bridgehead carbon. There's only one carbon
in our shortest path. So we put that in there. And then, there were 8 total
carbons, so it is octane. So the final IUPAC name
for this molecule is 6,8-dimethylbicyclo[3.2.1.]octane.